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alexandr1967 [171]
2 years ago
9

A 2. 0 kg ball is dropped from a rest 10m above the ground. How fast will the ball hit the ground

Physics
1 answer:
prohojiy [21]2 years ago
8 0

Answer:

A little over a second

Explanation:

gravity pulls down on objects at a speed of 9.8 meters a second

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Answer:

The heat causes the molecules on rubbing surfaces to move faster and have more energy.

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A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou
koban [17]

Answer:

The value is  f  =  0.05 \ Hz

Explanation:

From the question we are told that

     The mass of the car is  m  =  615 \  kg

      The period of the circular motion is  T  =  20 \  s

      The radius  is r =  80 \  m/s

Generally the frequency of the circular motion is  

       f  =   \frac{1}{T }

=>    f  =   \frac{1}{ 20  }

=>    f  =  0.05 \ Hz

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3 years ago
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3 years ago
Read 2 more answers
In an isolated system, Bicycle 1 and Bicycle 2, each with a mass of 10 kg,
dimulka [17.4K]

Answer: 20 kgm/s

Explanation:

Given that M1 = M2 = 10kg

V1 = 5 m/s , V2 = 3 m/s

Since momentum is a vector quantity, the direction of the two object will be taken into consideration.

The magnitude of their combined

momentum before the crash will be:

M1V1 - M2V2

Substitute all the parameters into the formula

10 × 5 - 10 × 3

50 - 30

20 kgm/s

Therefore, the magnitude of their combined momentum before the crash will be 20 kgm/s

7 0
3 years ago
You know your mass is 62 kg but when you stand on a bathroom scale in an elevator it says your mass is 77 kg what is the acceler
lbvjy [14]
In a stationary situation, the weight of person is
W=mg=(62 kg)(9.81 m/s^2) = 608.2 N
This is the weight "felt" by the scale, which is basically the normal reaction applied by the scale on the person, and which uses the value of g (9.81) as reference to convert the weight (602.8 N) into a mass (62 kg).

When the person is in the elevator, the scale says 77 kg. The scale is still using the same value of conversion (9.81), so the apparent weight "felt" by the scale is
W' = m'g=(77 kg)(9.81 m/s^2)=755.4 N
This is the normal reaction applied by the scale on the person, and which is directed upward. Besides this force, there is still the weight W of the person, acting downward. So, if we use Newton's second law:
\sum F = ma
W-W'=ma
where a is the acceleration of the elevator. If we solve for a, we find
a= \frac{W-W'}{m}= \frac{608.2N-755.4N}{62 kg}=-2.37 m/s^2
The negative sign means the acceleration is in the opposite direction of g (which we take positive), so it means the elevator is going upward.
4 0
3 years ago
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